A person walks 25.0 degrees north of east for 3.10 km. How far due north and how

Question

A person walks 25.0 degrees north of east for 3.10 km. How far due north and how far due east would she have to walk to arrive at the same location?

Explanation / Answer

(r, theta) --> (3.10 km, 25 degrees) [r = 3.10 km; theta = 25 degrees] x = rcostheta y = rsintheta x, in this case, would be the horizontal, or eastward distance, so the appropriate values would be substituted in the x equation to solve for it. x = 3.10 * (cos25) = 2.81 km y would be the vertical, or northward distance, so the appropriate values would be substituted in the y equation to solve for it. y = 3.10 * (sin25) = 1.31 km To verify the solutions, here is the optional step: (just to be sure!) r = sqrt (x^2 + y^2) r = sqrt (2.81^2 + 1.31^2) r = 3.0999 --> 3.10 km Therefore, the person must walk 1.31 km due north and 2.81 km due east to arrive at the same location.

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